Beginning at time t=0 we begin using bulbs, one at a time, to illuminate a room. Bulbs are replaced immediately upon failure. Each new bulb is selected independently by an equally likely choice between a Type-A bulb and a Type-B bulb.
The lifetime, X, of any particular bulb of a particular type is an independent random variable with the following PDF:
Why is the variance and expected value of the time before the first bulb failure equal to 2/3? Any idea of how to get this value? I got this and the answer from https://inst.eecs.berkeley.edu/~ee126/fa07/disc/week6hints.pdf .
$$\begin{align} \mathsf{E}(X)&=\mathsf{E}(X|A)\Pr(A)+\mathsf{E}(X|B)\Pr(B)\\ &=(1)(\frac{1}{2})+(\frac{1}{3})(\frac{1}{2})\\ &=\frac{2}{3} \end{align}$$
$$\begin{align} \mathsf{E}(X^2)&=\mathsf{E}(X^2|A)\Pr(A)+\mathsf{E}(X^2|B)\Pr(B)\\ &=2(\frac{1}{2})+(\frac{2}{9})(\frac{1}{2})\\ &=\frac{10}{9} \end{align}$$
$$\begin{align} \mathsf{Var}(X)&=\mathsf{E}(X^2)-\mathsf{E}^2(X)\\ &=\frac{10}{9}-(\frac{2}{3})^2\\ &=\frac{2}{3} \end{align}$$